- #1
math4everyone
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Homework Statement
Homework Equations
The z component of the field:
$$E_z = \frac{-Qh}{2\pi\varepsilon_0 (r^2+h^2)^{\frac{3}{2}}}$$
The Attempt at a Solution
I tried to choose a cylinder for my Gaussian surface such that the radius of it matches with the distance d I am trying to find and its height is such that all field lines are within the interior of the cylinder.
I know that the electric field in the z component in this case is:
\begin{equation}
\frac{-Qh}{2 \pi \varepsilon_o (r^2+h^2)^{\frac{3}{2}}}
\end{equation}
Thus, if I integrate:
\begin{equation}
\int \frac{-Qh}{2 \pi \varepsilon_o (r^2+h^2)^{\frac{3}{2}}}dA=\int_0^d \frac{-Qh(2 \pi r dr)}{2 \pi \varepsilon_o (r^2+h^2)^{\frac{3}{2}}}=\int_0^d \frac{-Qh(r dr)}{\varepsilon_o (r^2+h^2)^{\frac{3}{2}}}=\frac{Q}{\varepsilon_o}
\end{equation}
So:
\begin{equation}
\int_0^d \frac{r dr}{(r^2+h^2)^{\frac{3}{2}}}=-\frac{1}{h}
\end{equation}
And this is ultimately:
\begin{equation}
\frac{1}{h}-\frac{1}{\sqrt{h^2+d^2}}=-\frac{1}{h}
\end{equation}
However, if a solve for d, it yields a complex number
